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A184395
a(n) = number of distinct values obtained when sigma is applied to the divisors of n.
5
1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 6, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 10, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 7, 4, 7, 2, 6, 4, 8, 2, 12, 2, 4, 6, 6, 4, 8, 2, 10, 5, 4, 2, 12, 4, 4, 4, 8, 2, 12, 4, 6, 4, 4, 4, 12, 2, 6, 6, 9, 2, 8, 2, 8, 8, 4, 2, 12, 2, 8, 4, 10, 2, 8, 4
OFFSET
1,2
COMMENTS
Sequence is not the same as A000005(n): a(66) = 7, A000005(66) = 8.
a(n) = number of numbers k <= sigma(n) such that k = sigma(d) for some divisor d of n, where sigma = A000203. - This is the original name of the sequence, except that I substituted "some divisor" for "any divisor". - Antti Karttunen, Aug 24 2017
LINKS
FORMULA
a(n) = A000203(n) - A184396(n).
EXAMPLE
For n = 4, sigma(4) = 7, from numbers 1 - 7 there are three numbers k such that k = sigma(d) for any divisor d of n: 1 = sigma(1), 3 = sigma(2), 7 = sigma(4); a(4) = 3.
From Antti Karttunen, Aug 24 2017: (Start)
For n = 66, its 8 divisors are [1, 2, 3, 6, 11, 22, 33, 66]. When applying sigma to these, we obtain [1, 3, 4, 12, 12, 36, 48, 144], with one duplicate present, thus there are only 8-1 = 7 distinct values and a(66) = 7.
For n = 70, its 8 divisors are [1, 2, 5, 7, 10, 14, 35, 70]. When applying sigma to these, we obtain [1, 3, 6, 8, 18, 24, 48, 144], which are all unique values, thus a(70) = 8.
(End)
MATHEMATICA
Table[Length[Union[DivisorSigma[1, Divisors[n]]]], {n, 120}] (* Harvey P. Dale, Jun 20 2023 *)
PROG
(PARI) A184395(n) = length(vecsort(apply(d->sigma(d), divisors(n)), , 8)); \\ Antti Karttunen, Aug 24 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Jan 12 2011
EXTENSIONS
Name changed, a(66) and a(70) corrected and more terms added by Antti Karttunen, Aug 24 2017
STATUS
approved